A second-order system without zeros has its poles at s = −σ ± jωd. Which of the following correctly describes how the step response changes when only ωd is increased while σ is kept the same?

Answer: B. The settling time stays about the same, while the overshoot becomes larger and the period of oscillation shorter

The oscillation of the response decays within an envelope e^(−σt), so the settling time is set at roughly 4/σ and stays about the same if σ is unchanged. The overshoot is exp(−πσ/ωd), which grows as ωd increases. For example, raising ωd from 2 to 4 with σ = 2 increases the overshoot from about 4.3% to about 21% and shortens the period of oscillation 2π/ωd from about 3.1 s to about 1.6 s. If θ is the angle between the negative real axis and the line from the origin to the pole in the s-plane, then ζ = cos θ, and the larger θ is, the smaller the damping ratio.

• A. The overshoot depends on the ratio of σ to ωd, so changing ωd changes it.
• B. Correct. The rate of decay is set by the real part of the poles and the speed of oscillation by the imaginary part.
• C. A larger ωd lowers the damping ratio, so the overshoot becomes larger.